# Loop-the-loop problem help!

1. Nov 27, 2003

### Juntao

I'm going crazy over this problem. I've spent so much time on it that it sux.

I've included a picture for your pleasure.

A small spherical ball of radius r = 2.2 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 2.7 m. The ball has mass M = 378 g.
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a) How high above the top of the loop must it be released in order that the ball just makes it around the loop?
b) Repeat problem (a) for a disk. Find the ratio of the heights h for the two cases.

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Okay,I'm just stuck on part a. If I figure that out, rest should be peice of cake.
I know that I'd have to use conservation of energy. The ball starts off with potential energy, and that converts into translational and rotational energy.

So the equation that I've set up is:
mgh=.5MV^2+.5*I*w^2 (equation 1)

where I is moment of inertia
and w is omega.
Since it is a sphere, the I= 2/5Mr^2

So right now I know
m=.378kg
g=9.8m/s/s
r=.022m
but not omega or velocity, but I can calculate those two.

At the top of the loop, there is a normal force and gravity force acting on the ball. So thats 2*mg, or 7.40N.
Centripetal force = m(V^2/R)
Solving that, I get velocity equals 7.27 m/s
Then v=w*r
Using that velocity number, I get omega=330.45/s

So I plugged in all my values into equation 1, solve for H, and I get H=1.43m, but that isn't right. How do I solve this? I think I'm on the right track, but something along the way is screwing me over!

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Last edited: Nov 27, 2003
2. Nov 27, 2003

### arcnets

Juntao,
I think your method is basically correct, except for the following errors:
I think the force of gravity is just mg, not 2mg.
I think, since the ball rolls on the inside, the actual radius is R-r, so
$$\textit{centripetal force} = \frac {mv^2}{R-r}$$
Solve again for v and &omega;, and proceed as you did.
How can you see this isn't right? I think the correct answer is in the same order of magnitude.

Last edited: Nov 27, 2003
3. Nov 27, 2003

### Staff: Mentor

Re: Re: loop-the-loop problem help!

Juntao: arcnets has pointed out your main error; do as he suggests!

1) Don't plug numbers (like for m, g, etc.) in immediately; instead, do as much of your work algebraically until the last step. Only plug numbers in at the end. You'll find that lots of things will cancel and that calculations will be easier and less prone to error.

2) Don't forget that you are to find the height above the loop. When the sphere is at the top of its path in the loop, it's really below the level of the top of the loop.

4. Nov 27, 2003

### Juntao

Still figuring this thing out, but I was wondering, when the sphere is at the top of the loop, why isn't there a normal force? So at the top, it only has gravitational force?

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Ok, I'm utilizing advice given by arcnets and Doc Al and this is what happens.

centripetal force
mg=(m*V^2)/(R-r)

or v=sqrt[g(R-r)]

I know this equation v=w*r
so w= (sqrt[g(R-r)])/r

Using the conservation of energy:
mgh=.5*m*V^2+.5*I*w^2
goes to
mgh=.5*m(sqrt[g(R-r)])^2+.5[2/5*m*r^2]+((sqrt[g(R-r)])/r)^2

when I simplify all of that, in the end I get:
h=(7/10)(R-r)
so h equal= (7/10)(2.7m-.022m)=1.8746m. My god, it worked! Thank you guys so much, and takes for the tip Doc Al about using algebra. It did make my life simpler.

5. Nov 27, 2003

### HallsofIvy

Staff Emeritus
There IS a normal force. What does "normal" mean in "normal force"?

6. Nov 27, 2003

### Sniper__1

are you assuming no friction and if not what is the loop and ramp made of and what is that materialsfriction against the ball and the same thing with the ball.

7. Nov 28, 2003

### Staff: Mentor

Let's be clear about terminology. At the top of its path, the sphere has two forces acting on it. Its weight and the "normal force", which is the force that the loop surface exerts on the ball.

This normal force is, in a sense, a passive force: it will be whatever it needs to be to contain the sphere. (Until the loop breaks that is!) In this problem, you are trying to find the minimum energy to have the sphere just make it around the loop. That minimum is when the normal force equals zero. (Any less energy, and it won't make it to the top of the loop.)

8. Nov 28, 2003

### arcnets

Juntao,
one more detail: The h that you calculated ...
$$h = \frac{7}{10}(R-r)$$
is the (IMO correct) height difference of the ball's center-of-mass in final and initial state.
Whereas the h in the picture is taken from top-of-loop to initial center-of-mass. So:
$$h = \frac{7}{10}(R-r) - r = \frac{7}{10}R-\frac{17}{10}r$$